YES(O(1),O(n^2))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { or(x, true()) -> true()
  , or(true(), y) -> true()
  , or(false(), false()) -> false()
  , mem(x, nil()) -> false()
  , mem(x, set(y)) -> =(x, y)
  , mem(x, union(y, z)) -> or(mem(x, y), mem(x, z)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.

Trs:
  { or(x, true()) -> true()
  , or(true(), y) -> true()
  , or(false(), false()) -> false()
  , mem(x, union(y, z)) -> or(mem(x, y), mem(x, z)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
     [or](x1, x2) = 1 + x1 + x2      
                                     
         [true]() = 0                
                                     
        [false]() = 0                
                                     
    [mem](x1, x2) = x1 + x1*x2 + 2*x2
                                     
          [nil]() = 0                
                                     
        [set](x1) = x1               
                                     
      [=](x1, x2) = x1 + x2          
                                     
  [union](x1, x2) = 1 + x1 + x2      
                                     
  
  This order satisfies the following ordering constraints.
  
           [or(x, true())] =  1 + x                          
                           >                                 
                           =  [true()]                       
                                                             
           [or(true(), y)] =  1 + y                          
                           >                                 
                           =  [true()]                       
                                                             
    [or(false(), false())] =  1                              
                           >                                 
                           =  [false()]                      
                                                             
           [mem(x, nil())] =  x                              
                           >=                                
                           =  [false()]                      
                                                             
          [mem(x, set(y))] =  x + x*y + 2*y                  
                           >= x + y                          
                           =  [=(x, y)]                      
                                                             
     [mem(x, union(y, z))] =  2*x + x*y + x*z + 2 + 2*y + 2*z
                           >  1 + 2*x + x*y + 2*y + x*z + 2*z
                           =  [or(mem(x, y), mem(x, z))]     
                                                             

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { mem(x, nil()) -> false()
  , mem(x, set(y)) -> =(x, y) }
Weak Trs:
  { or(x, true()) -> true()
  , or(true(), y) -> true()
  , or(false(), false()) -> false()
  , mem(x, union(y, z)) -> or(mem(x, y), mem(x, z)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.

Trs: { mem(x, set(y)) -> =(x, y) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
     [or](x1, x2) = x1 + x2            
                                       
         [true]() = 0                  
                                       
        [false]() = 0                  
                                       
    [mem](x1, x2) = 2*x1 + 2*x1*x2 + x2
                                       
          [nil]() = 0                  
                                       
        [set](x1) = 1 + x1             
                                       
      [=](x1, x2) = x1 + x2            
                                       
  [union](x1, x2) = 2 + x1 + x2        
                                       
  
  This order satisfies the following ordering constraints.
  
           [or(x, true())] =  x                              
                           >=                                
                           =  [true()]                       
                                                             
           [or(true(), y)] =  y                              
                           >=                                
                           =  [true()]                       
                                                             
    [or(false(), false())] =                                 
                           >=                                
                           =  [false()]                      
                                                             
           [mem(x, nil())] =  2*x                            
                           >=                                
                           =  [false()]                      
                                                             
          [mem(x, set(y))] =  4*x + 2*x*y + 1 + y            
                           >  x + y                          
                           =  [=(x, y)]                      
                                                             
     [mem(x, union(y, z))] =  6*x + 2*x*y + 2*x*z + 2 + y + z
                           >  4*x + 2*x*y + y + 2*x*z + z    
                           =  [or(mem(x, y), mem(x, z))]     
                                                             

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs: { mem(x, nil()) -> false() }
Weak Trs:
  { or(x, true()) -> true()
  , or(true(), y) -> true()
  , or(false(), false()) -> false()
  , mem(x, set(y)) -> =(x, y)
  , mem(x, union(y, z)) -> or(mem(x, y), mem(x, z)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.

Trs: { mem(x, nil()) -> false() }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
     [or](x1, x2) = x1 + x2              
                                         
         [true]() = 0                    
                                         
        [false]() = 0                    
                                         
    [mem](x1, x2) = 1 + x1 + x1*x2 + 2*x2
                                         
          [nil]() = 0                    
                                         
        [set](x1) = x1                   
                                         
      [=](x1, x2) = x1 + x2              
                                         
  [union](x1, x2) = 2 + x1 + x2          
                                         
  
  This order satisfies the following ordering constraints.
  
           [or(x, true())] =  x                              
                           >=                                
                           =  [true()]                       
                                                             
           [or(true(), y)] =  y                              
                           >=                                
                           =  [true()]                       
                                                             
    [or(false(), false())] =                                 
                           >=                                
                           =  [false()]                      
                                                             
           [mem(x, nil())] =  1 + x                          
                           >                                 
                           =  [false()]                      
                                                             
          [mem(x, set(y))] =  1 + x + x*y + 2*y              
                           >  x + y                          
                           =  [=(x, y)]                      
                                                             
     [mem(x, union(y, z))] =  5 + 3*x + x*y + x*z + 2*y + 2*z
                           >  2 + 2*x + x*y + 2*y + x*z + 2*z
                           =  [or(mem(x, y), mem(x, z))]     
                                                             

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

Weak Trs:
  { or(x, true()) -> true()
  , or(true(), y) -> true()
  , or(false(), false()) -> false()
  , mem(x, nil()) -> false()
  , mem(x, set(y)) -> =(x, y)
  , mem(x, union(y, z)) -> or(mem(x, y), mem(x, z)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

Hurray, we answered YES(O(1),O(n^2))